6.003 Homework #12 Solutions
Problems
1. Which are True?
For each of the DT signals x1 [n] through x4 [n] (below), determine whether the conditions
listed in the following table are satisfied, and answer T for true or F for false.
x1 [n]
x2 [n]
x3 [n]
x4 [n]
T
F
T
T
T
T
T
F
X(e jΩ ) is purely imaginary
F
F
T
F
e jkΩ X(e jΩ ) is purely real for some integer k
F
T
F
F
X(e j0 ) = 0
Rπ
−π X(e
jΩ ) dΩ
=0
x1 [n]
1
n
x2 [n]
1
n
x3 [n]
1
n
x4 [n]= x4 [n + 5]
1
n
6.003 Homework #12 Solutions / Fall 2011
2
2. Inverse Fourier
The magnitude and angle of the Fourier transform of x[n] are shown below.
magnitude
1
| sin Ω
2|
Ω
−2π
0
2π
angle
π
π/2
Ω
−2π
2π
−π/2
−π
Determine x[n].
x[n]
1
2
n
1
− 12
X(ejΩ ) = −j sin
1
x[n] =
2π
Ω −jΩ/2
e
= −j
2
�
X(e
2π
⎧
⎨ −1/2
= 1/2
⎩
0
jΩ
)e
jΩn
e jΩ/2 − e−jΩ/2
j2
1
dΩ =
4π
n=0
n=1
otherwise
�
1
1
e−jΩ/2 = e−jΩ −
2
2
�
� −jΩ
1
− 1 e jΩn dΩ =
e
4π
2π
�
2π
�
�
e jΩ(n−1) − e jΩn dΩ
3
6.003 Homework #12 Solutions / Fall 2011
Engineering Design Problem
3. Sampling with alternating impulses
A CT signal xc (t) is converted to a DT signal xd [n] as follows:
xd [n] =
xc (nT )
n even
−xc (nT ) n odd
a. Assume that the Fourier transform of xc (t) is Xc (jω ) shown below.
Xc (jω)
1
ω
W
Determine the DT Fourier transform Xd (ejΩ ) of xd [n].
Let xe [n] = (−1)n xd [n] = xc (nT ). Then
Z
Z
1
n
n 1
j Ω jΩn
Xd (ejΩ )e j Ωn dΩ
xe [n] =
Xe (e )e
dΩ = (−1) xd [n] = (−1)
2π 2π
2π 2π
Z
Z
1
1
= e−jnπ
Xd (ej Ω )e jΩn dΩ =
Xd (ej Ω )e j (Ω−π)n dΩ
2π 2π
2π 2π
Z
1
0
0
=
Xd (ej(Ω +π) )e jΩ n dΩ0
2π 2π
Thus Xe (e jΩ ) = Xd (e j(Ω+π) ). Also
Z
Z ∞
1
1
j Ω jΩn
xe [n] =
Xe (e )e
dΩ = xc (nT ) =
Xc (jω)e jωnT dω
2π 2π
2π −∞
jΩ
Substituting ω = Ω
T we see that Xe (e ) =
|ω| > Tπ . It follows that
Xd (e jΩ ) =
1
Ω
T Xc (jω)|ω= T
provided Xc (jω) = 0 for
1
Xc (jω)|ω= Ω−π
T
T
as shown in the following figure.
Xd (e jΩ )
1
T
Ω
π−W T π π+W T
An alternative solution is to construct a novel “sampling function” that consists of
alternating impulses (hence the title of the problem).
pa (t) =
∞
X
(−1)n δ (t − nT )
n=−∞
You can think of this signal as two times an impulse train with period 2T minus an
impulse train with period T
pa (t) =
∞
X
n=−∞
2δ(t − 2nT ) −
∞
X
n=−∞
δ(t − nT )
6.003 Homework #12 Solutions / Fall 2011
4
The Fourier transform is then
∞
∞
X
X
2π
2π n
2π
2πn
Pa (jω) =
2
δ ω−
−
δ ω−
2T
2T
T
T
n=−∞
n=−∞
∞
X
=
n = −∞
n odd
2π πn δ ω−
T
T
which is shown below
Pa (jω)
2π
T
ω
− 3π
T
− Tπ
π
T
3π
T
Convolving this function of frequency with Xc (jω) and converting to the DTFT leads
jΩ ) that was shown above.
to the same solution for Xd (ejΩ
b. Assume that xc (t) is bandlimited to −W ≤ ω ≤ W . Determine the maximum value
of W for which the original signal xc (t) can be reconstructed from the samples xd [n].
From the previous part, π − W T must be greater than 0. Thus
π
W < .
T
6.003 Homework #12 Solutions / Fall 2011
5
4. Boxcar sampling
A digital camera focuses light from the environment onto an imaging chip that converts
the incident image into a discrete representation composed of pixels. Each pixel represents
the total light collected from a region of space
� mD+ ∆ Z nD+ ∆
Z
2
2
xc (x, y) dx dy
xd [n, m] =
mD− ∆
2
nD− ∆
2
where Δ is a large fraction of the distance D between pixels. This kind of sampling is
often called “boxcar” sampling to distinquish it from the ideal “impulse” sampling that
we described in lecture. Assume that boxcar sampling is defined in one dimension as
� nT + ∆
Z
2
xc (t) dt
xd [n] =
nT − ∆
2
where T is the intersample “time.”
a. Let Xc (jω) represent the continuous-time Fourier transform of xc (t). Determine the
discrete-time Fourier transform Xd (e jΩ ) of xd [n] in terms of Xc (jω), Δ, and T .
If we define
p(t) =
1 −∆
2 <t<
0 otherwise
∆
2
then we can express xd [n] as xb (nT ) where xb (t) = (xc ∗ p)(t). The Fourier transform
of xb (t) is then the product of the Fourier transforms of xc (t) and p(t).
Z ∆
2
2 sin ω∆
2
P (jω) =
e−jωt dt =
∆
ω
−
2
Provided that Xb (jω) is 0 for |ω| >
π
T,
2 sin Ω∆
1
1
Ω
2T
Xd (e ) = Xb (jω )|ω= Ω = (Xc (jω )P (jω )) |ω= Ω =
Xc j
T
T
T
T
Ω
T
jΩ
over the interval −π < Ω < π and repeats periodically (with period 2π) outside that
interval.
b. Assume that xc (t) is bandlimited to −W ≤ ω ≤ W . Determine the the maximum
value of W for which the original signal xc (t) can be reconstructed from the samples
xd [n]. Compare your answer to the answer for an ideal “impulse” sampler.
Since Xb (jω) must be 0 for |ω| > Tπ (see above), it follows that Xc (jω) must be zero
over this same range. Therefore, Xc must be bandlimited in W = Tπ . Furthermore,
P (jω) is non-zero for |ω| < π/T , so all of the information in Xc can be reconstructed
from that in Xb (i.e., there is no loss of information in going from Xc to Xb ). Thus
boxcar sampling has exactly the same frequency limitations as impulse sampling.
6.003 Homework #12 Solutions / Fall 2011
6
c. Describe the effect of boxcar sampling on the resulting samples xd [n]. How are the
samples that result from boxcar sampling different from those that result from impulse
sampling?
The samples generated by boxcar sampling are lowpass filtered relative to those gener­
ated by impulse sampling. The attenuation is greatest ( π2 ) at the maximum frequency
Ω = π and for the largest possible value of Δ, which is T . The attenuation is less for
lower frequencies and for smaller values of Δ.
7
6.003 Homework #12 Solutions / Fall 2011
5. DT processing of CT signals
Sampling and reconstruction allow us to process CT signals using digital electronics as
shown in the following figure.
xc (t)
impulse
sampler
xd [n]
yd [n]
hd [n]
impulse
reconstruction
yp (t)
ideal
LPF
yc (t)
The “impulse sampler” and “impulse reconstuction” use sampling interval T = π/100.
The unit-sample function hd [n] represents the unit-sample response of an ideal DT low­
pass filter with gain of 1 for frequencies in the range − π2 < Ω < π2 . The “ideal LPF”
passes frequencies in the range −100 < ω < 100. It also has a gain of T throughout its
pass band.
Assume that the Fourier transform of the input xc (t) is X(jω) shown below.
Xc (jω)
1
ω
−100
100
Determine Yc (jω).
Impulse sampling of xc (t) produces xd [n] = xc (nT ). Substituting into the transform
relations shows that
Z ∞
Z
1
1
jΩ j Ωn
xd [n] =
Xd (e )e
dΩ = xc (nT ) =
Xc (jω)e jωnT dω
2π 2π
2π −∞
Thus
Xd (e jΩ ) =
since dω =
1
Xc (jω)|ω= Ω
T
T
1
T dΩ.
Therefore
1
Ω
100
Ω
jΩ
Xd (e ) = Xc j
=
Xc j
.
T
T
π
π/100
π
Notice that the cutoff frequency ω = 100 maps to Ω = ωT = 100 × 100
= π. Also, Xd (e j Ω )
is periodic in 2π as are all DT Fourier transforms, as shown below.
Xd (e jΩ )
100
π
ω
π
−2π −π
2π
The output of the ideal lowpass filter has the following transform.
Yd (e j Ω )
100
π
ω
−2π −π
π
2π
Impulse reconstruction then generates a signal with the following transform.
6.003 Homework #12 Solutions / Fall 2011
8
Yp (jΩ)
100
π
ω
−200 −100
100
200
The final output has the following transform, where the DC value is
100
π
× T = 1.
Yc (jΩ)
1
1
2
ω
−50
50
The overall effect is that the input signal is lowpass filtered by the discrete system.
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6.003 Signals and Systems
Fall 2011
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